Translation-Equivariant Rules In A Gaussian Location Model

Why Translation Equivariance Matters

In a Gaussian location problem, changing the unknown location by an amount $t$ shifts the distribution of the observation by the same amount. A rule that respects this structure should shift its estimate by $t$ as well. This property, called translation equivariance, places a strong restriction on what an estimator can look like when there is only one observation.

The restriction is especially useful because it reduces an entire class of rules to a one-parameter family. Once the fixed shift is known, its bias, variance, and squared-error risk follow directly. This is a characterization of equivariant rules; by itself, it does not claim that every such rule is best among all possible estimators.

Model and definition: Let $X\sim N(\theta,\sigma^2)$, where $\theta\in\mathbb{R}$ is unknown and $\sigma^2>0$ is known. A nonrandomized estimator is a measurable function $\delta:\mathbb{R}\to\mathbb{R}$. It is translation-equivariant if, for every observation $x\in\mathbb{R}$ and every translation $t\in\mathbb{R}$, $$ \delta(x+t)=\delta(x)+t. $$ Under squared-error loss, its risk at $\theta$ is $R(\theta,\delta)=E_\theta[(\delta(X)-\theta)^2]$, where $E_\theta$ denotes expectation when $X\sim N(\theta,\sigma^2)$.

The definition applies to the rule as a function of every possible observation, not just to its average behavior under one particular value of $\theta$. The same translation $t$ is applied to the data and to the location parameter. Thus the condition expresses that the rule has no preferred origin on the real line.

Worked Examples Before the General Result

Worked Example: A Fixed Downward Shift

Suppose $X\sim N(\theta,9)$, and consider $\delta(x)=x-2$. Check equivariance and calculate its squared-error risk for an arbitrary $\theta$.

For any $x,t\in\mathbb{R}$, $$ \delta(x+t)=(x+t)-2=(x-2)+t=\delta(x)+t, $$ so the rule is translation-equivariant. Its estimation error is $$ \delta(X)-\theta=(X-2)-\theta=(X-\theta)-2. $$ Since $E_\theta[X-\theta]=0$ and $\operatorname{Var}_\theta(X-\theta)=9$, the error has bias $-2$ and variance $9$. Therefore, $$ R(\theta,\delta)=9+(-2)^2=13. $$ The answer does not depend on $\theta$; the rule’s constant downward shift contributes $4$ to risk, in addition to the observation variance $9$.

Worked Example: Comparing Two Equivariant Shifts

Let $X\sim N(\theta,4)$, and compare $\delta_1(x)=x-1$ with $\delta_2(x)=x+0.5$. Find each rule’s bias and risk for every $\theta$.

Both rules have the form $x+c$ for a fixed constant $c$, so translating $x$ by $t$ translates either estimate by exactly $t$. For $\delta_1$, the error is $(X-\theta)-1$, which has mean $-1$ and variance $4$. Its risk is $4+1^2=5$. For $\delta_2$, the error is $(X-\theta)+0.5$, which has mean $0.5$ and variance $4$. Its risk is $4+0.5^2=4.25$.

Thus $R(\theta,\delta_1)=5$ and $R(\theta,\delta_2)=4.25$ for every $\theta$. The second rule has lower risk in this comparison because its shift is closer to zero. Neither calculation depends on guessing the unknown location.

Worked Example: A Shrinkage Rule That Is Not Equivariant

Suppose $X\sim N(\theta,1)$ and consider $\delta(x)=0.8x$. Determine whether it is translation-equivariant, then calculate and compare its risk at $\theta=0$ and $\theta=5$.

For a translation $t$, $\delta(x+t)=0.8x+0.8t$, whereas $\delta(x)+t=0.8x+t$. These are not equal for general $t$, so the rule is not translation-equivariant. Its error can be written as $$ \delta(X)-\theta=0.8(X-\theta)-0.2\theta. $$ The first term has mean zero and variance $0.8^2=0.64$; the second term is constant. Consequently the risk is $0.64+(-0.2\theta)^2=0.64+0.04\theta^2$. At $\theta=0$, the risk is $0.64$. At $\theta=5$, it is $0.64+0.04(25)=1.64$.

The risk changes with $\theta$, unlike the constant risk of a translation-equivariant rule. This example also shows why one should not infer equivariance merely because an estimator is a simple linear function of the observation.

Characterizing the Equivariant Rules

The key fact is elementary but powerful: the equivariance condition can be evaluated at the observation $0$. It then determines the rule at every other point. No differentiability, continuity, or linearity assumption is needed.

Theorem: A function $\delta:\mathbb{R}\to\mathbb{R}$ is translation-equivariant if and only if there is a constant $c\in\mathbb{R}$ such that $$ \delta(x)=x+c \qquad\text{for every }x\in\mathbb{R}. $$

Proof. Suppose first that $\delta$ is translation-equivariant. Set $x=0$ and $t=u$ in the defining condition. For every $u\in\mathbb{R}$, $$ \delta(u)=\delta(0)+u. $$ Taking $c=\delta(0)$ gives $\delta(u)=u+c$ for every $u$. Conversely, if $\delta(x)=x+c$ for a fixed $c$, then for every $x,t$, $$ \delta(x+t)=(x+t)+c=(x+c)+t=\delta(x)+t. $$ Thus the rule is translation-equivariant. The two implications establish the characterization. $\square$

The constant $c$ is the rule’s shift relative to the observation. It is a feature of the estimator, not a value that changes with the data or with the unknown parameter. The characterization includes the unshifted observation $X$ when $c=0$, as well as every fixed upward or downward shift.

Deriving the Constant Risk

The characterization gives a direct risk calculation. Write $X=\theta+(X-\theta)$. Under the model, $X-\theta\sim N(0,\sigma^2)$, so it has mean zero and variance $\sigma^2$. For the rule $\delta(X)=X+c$, the estimation error is $$ \delta(X)-\theta=(X-\theta)+c. $$ Its mean is $c$, and its variance is $\sigma^2$. Using the squared-error decomposition into variance plus squared bias gives the following result.

Theorem: In the model $X\sim N(\theta,\sigma^2)$, with $\sigma^2>0$ known and $\theta\in\mathbb{R}$, the translation-equivariant rule $\delta(X)=X+c$ has bias $c$, variance $\sigma^2$, and risk $$ R(\theta,\delta)=\sigma^2+c^2 $$ for every $\theta\in\mathbb{R}$.

Proof. The error is $X-\theta+c$. Since $E_\theta[X-\theta]=0$, its expectation is $c$. Adding a constant does not change variance, so $$ \operatorname{Var}_\theta(\delta(X)-\theta) =\operatorname{Var}_\theta(X-\theta+c) =\operatorname{Var}_\theta(X-\theta) =\sigma^2. $$ The error has a finite second moment because it is a normal random variable shifted by a finite constant. Therefore the bias-variance identity applies, yielding $$ R(\theta,\delta) =\operatorname{Var}_\theta(\delta(X)-\theta) +\big(E_\theta[\delta(X)-\theta]\big)^2 =\sigma^2+c^2. $$ Neither term depends on $\theta$, proving the stated constant-risk formula. $\square$

There is also a direct check by expanding the square. The cross term has expectation zero because $E_\theta[X-\theta]=0$: $$ E_\theta[((X-\theta)+c)^2] =E_\theta[(X-\theta)^2]+2cE_\theta[X-\theta]+c^2 =\sigma^2+c^2. $$ This calculation verifies the same formula without relying on any uncomputed covariance or cross term.

What the Result Does—and Does Not—Say

Within the family of translation-equivariant rules, the risk is minimized by $c=0$. Indeed, for every real $c$, $c^2\geq 0$, with equality exactly when $c=0$. The observation rule $\delta(X)=X$ therefore has risk $\sigma^2$, and every nonzero fixed shift has strictly larger risk $\sigma^2+c^2$. This is a comparison within the characterized family; the calculation alone does not establish a general optimality claim over every estimator.

A constant risk is not the same thing as a small risk. The rule $X+100$ is translation-equivariant and has constant risk $\sigma^2+10{,}000$. Equivariance describes how a rule responds to a change in location; it does not impose a useful size on its fixed offset. Similarly, the lower risk of one shifted rule in a particular comparison follows from the numerical values of its shift and variance, not from equivariance alone.

Common Pitfall: Confusing Scaling With Shifting

A tempting but incorrect argument is: “The estimator $aX$ is linear, so translating the data by $t$ must translate the estimate by $t$.” The actual change in the estimate is $a(X+t)-aX=at$, not $t$. Unless $a=1$, those amounts differ for nonzero translations. A linear estimator is not automatically translation-equivariant.

The correction is to apply the defining condition directly. For $\delta(x)=ax$, equivariance requires $a(x+t)=ax+t$ for every $x,t$, which reduces to $at=t$ for every $t$, and hence to $a=1$. This also explains why the characterization theorem does not conflict with rules such as $0.8X$: those rules are outside the equivariant class.

Key takeaway: For one real Gaussian observation with unknown location and known variance, exact translation equivariance forces the rule to be $X+c$ for a fixed constant $c$. Its bias is $c$, its variance is $\sigma^2$, and its squared-error risk is the location-independent value $\sigma^2+c^2$.

Check Your Understanding

Use the definition of translation equivariance and the risk formula to answer these questions.

Show directly that $\delta(x)=x+3$ is translation-equivariant, and give its bias and risk when $\sigma^2=2$.
If a rule is translation-equivariant and $\delta(0)=-1.5$, what is the rule’s value at every $x$?
For $X\sim N(\theta,5)$, calculate the risk of the equivariant rule $X-0.4$ for arbitrary $\theta$.
For which value of $a$ is the rule $\delta(x)=ax$ translation-equivariant? Verify using the definition.
Why does constant risk across $\theta$ not imply that every translation-equivariant rule has the same risk?